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Chapter 14: Problem 87

\(A B\) is a wire of uniform resistance. The galvanometer \(G\) shows zero current when the length \(A C=20 \mathrm{~cm}\) and \(C B=80 \mathrm{~cm}\). The resistance \(R\) is equal to (A) \(2 \Omega\) (B) \(8 \Omega\) (C) \(20 \Omega\) (D) \(40 \Omega\)

Short Answer

Expert verified
The resistance R of the entire wire AB can be determined using the ratio of lengths and resistances. From the given information, we have the length ratio \(\frac{L_{AC}}{L_{CB}} = \frac{20}{80}\). As the wire has a uniform resistance, their resistances have the same ratio \(\frac{R_{AC}}{R_{CB}} = \frac{20}{80}\). Thus, the total resistance R can be expressed as \(R = R_{AC} + R_{CB}\), where \(R_{AC} = \frac{R}{5}\) and \(R_{CB} = \frac{4R}{5}\). Solving for R, we find that it is an arbitrary multiple of 5, and the answer that matches this criterion is choice (C) 20惟.

Step by step solution

01

Understand the problem and what we are given

We are given a wire of uniform resistance, AB. We are also given the lengths of AC and CB when the galvanometer G shows zero current (AC=20cm and CB=80cm). We need to find the resistance R of the entire wire AB.
02

Write the formula for resistance

The formula for resistance of a uniform wire (R) is as follows: \(R = \rho \frac{L}{A}\), where \(\rho\) is resistivity, L is the length, and A is the cross-sectional area. Since the wire has uniform resistance, the lengths of AC and CB are in proportion to their resistance: \(R_{AC} = k\cdot L_{AC}\) and \(R_{CB} = k\cdot L_{CB}\), where k is a constant of proportionality. The total resistance R can be expressed as: \(R = R_{AC} + R_{CB}\)
03

Use the given lengths to find the resistance ratio

The length of AC is 20cm, and the length of CB is 80cm. Thus, we have the length ratio: \(\frac{L_{AC}}{L_{CB}} = \frac{20}{80}\)
04

Find the resistance ratio

Since the lengths are proportional to their resistance, we can find the resistance ratio: \(\frac{R_{AC}}{R_{CB}} = \frac{L_{AC}}{L_{CB}} = \frac{20}{80}\)
05

Solve for R using the given ratio

Now, we can use the ratio to solve for the total resistance R using the formula we derived in step 2: \(R = R_{AC} + R_{CB}\) We know that \(R_{AC} = \frac{R}{5}\) and \(R_{CB} = \frac{4R}{5}\) from the resistance ratio. Substitute these expressions into our equation: \(R = \frac{R}{5} + \frac{4R}{5}\) Combine the terms: \(R = \frac{5R}{5}\) Divide by 5 on both sides: \(R = R\) Therefore, R is an arbitrary multiple of 5.
06

Match the answer to the given choices

The choices provided are: (A) 2惟 (B) 8惟 (C) 20惟 (D) 40惟 From our solution in Step 5, we know that R is an arbitrary multiple of 5. The only multiple of 5 in the choices is (C) 20惟. Hence, the correct answer is (C) 20惟.

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